Trouble with Lorentz transformations

[starthaus]

We would not get a "good" calculation of x₂ - x₁ unless t₁ = t₂. If I calculate t₁ and t₂ from the data given on x₁', x₂', t₁' and t₂', I will NOT get t₁ = t₂ because of this relativity of simultaneity.

Right. So you can EITHER calculate L as L'/gamma OR L' as L/\gamma.

 

[pc2-brazil]

starthaus: Let's say I make the measurement of L from O and want to measure by which factor L is smaller than L', the proper length (length measured by O', according to which the measured object is at rest). Why can't I obtain this relation (which is L = L'/γ) by writing x'₂ - x'₁ with t₂ = t₁?
I still don't understand why this is wrong. This seems to be calculated like this very commonly. For example:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html

 

[starthaus]

starthaus: Let's say I make the measurement of L from O and want to measure by which factor L is smaller than L', the proper length (length measured by O', according to which the measured object is at rest). Why can't I obtain this relation (which is L = L'/γ) by writing x'₂ - x'₁ with t₂ = t₁?
I still don't understand why this is wrong.

Because the theory of measurement says that , in order to calculate the length of an objects in a frame, you need to mark its endpoints simultaneously in THAT particular frame.

This seems to be calculated like this very commonly. For example:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html

It says quite clearly, that it calculates the length x_2-x_1 for t_2=t_1, exactly what I have been saying several times in this thread.

 

[pc2-brazil]

It says quite clearly, that it calculates the length x_2-x_1 for t_2=t_1, exactly what I have been saying several times in this thread.

And also what I was trying to say.
If that in the site is written correctly, then what I wrote in the last phrase of post #14 is correct.
Unless I expressed myself badly. I want to measure x₂ - x₁ with t₂ = t₁, but start the resolution of the problem by writing x'₂ - x'₁ and substituting it for the Lorentz transformations instead of doing it the other way: writing x₂ - x₁. Then, I obtain L = L'/γ.

 

[starthaus]

And also what I was trying to say.
If that in the site is written correctly, then what I wrote in the last phrase of post #14 is correct.
Unless I expressed myself badly. I want to measure x₂ - x₁ with t₂ = t₁, but start the resolution of the problem by writing x'₂ - x'₁ and substituting it for the Lorentz transformations instead of doing it the other way: writing x₂ - x₁. Then, I obtain L = L'/γ.

ok, then

 

[stevmg]

Dr. Einstein, with very simple algebra and calculations proves what both of you (pc2-brazil and starthaus) are saying: L = L'/\gamma. In his book, Relativity, Section XII "The Behaviour of Measuring Rods and Clocks in Motion" he puts his derivation in this format as opposed to L' = \gammaL. If you assume that x₁' and x₂' are given in O' and x₁' and x₂' are fixed in O' and you "Lorentz" them to x₁ and x₂ in O and you do it in a way that pc2-brazil algebraically defines, you will wind up with t₁ = t₂ (because that's the way he set it up - to come out that way.) Thus, you are measuring x₁ and x₂ at he same time (simultaneously) in O. Because simultaneity is not preserved from different frames of reference, in O', the "corresponding t' s" are not equal to each other as they were in O. This is algebraically established in the second to last paragraph of this post. But that doesn't make any difference. Since x₁' and x₂' are fixed in O' as this was a given, the actual times of measurement of x₁' and x₂' in O' are immaterial to their measurements because these measurements never change in that frame of reference.

Final point, to establish that I do understand the derivation of L = L'/\gamma or L' = \gammaL:

x₂'= \gamma[x₂ - vt₂]
x₁'= \gamma[x₁ - vt₁]
x₂' - x₁' = L' = \gamma[x₂ - x₁ - (vt₂ - vt₁)]
but t₂ = t₁, therefore
x₂' - x₁' = L' = \gamma[x₂ - x₁ - (vt₂ - vt₂)] = \gamma(x₂ - x₁) = L
L' = \gammaL
Q.E.D.

Just for the record:
t₂' = \gamma(t₂ - vx₂)
t₁' = \gamma(t₁ - vx₁)
but t₂ = t₁, therefore
t₂' = \gamma(t₂ - vx₂)
t₁' = \gamma(t₂ - vx₁)
t₂' - t₁' = \gamma[v(x₁ - x₂)] which is not zero.
This shows that t₂' does not = t₁'. Again, this is irrelevant as x₁' and x₂' never change with time in O' because this was stipulated as a given so the L' = \gammaL relationship holds.

If I am wrong, please, please, please do NOT call me a dolt. I was born during the day, but not yesterday. Pardon any typos. I tried to clean them up but this is so damn cumbersome I may have missed a few.

 

[stevmg]

To starthaus:

These two quotes answers the "why" one can use x₁' and x₂' in which t₁' \neq t₂' (i.e, differing times when the rod is measured between front and back) for the calculation which t₁ = t₂ in S (or frame O in the pc2-brazil question.) Up until that point I couldn't make the "jump" and figure how you were still able to use the length of the rod in S' or L' as pc2-brazil stated it with its differing times of measurement of the front and back to calculate the length (L) in frame O in which the times of measurement of the front and back were the same.

Back to Frame-Dragger:

If I do place a rod of certain length l₀ in the moving frame S' which moves at a velocity of v to the right with respect to the stationary frame S and measure it in the stationary frame S at x₁ and x₂ simultaneously at t₁ = t₂, we get the length l₀ = x₂ - x₁ according to the length contraction formula l₀ = l/gamma. However, when we solve the simultaneous Lorentz equations for t_1' and t₂' in in S' which are the corresponding times for the t₁ = t₂ in S we note that t₁' does not = t₂\neq' which means we measured the front and back of the rod at different times in S'.

Since the rod was stationary in S' it does not make a difference for l = x₂' - x₁' as both x₂' and x₁' are fixed in S' so that l will always be the same no matter when you measure the rod even if you measure the front and back at different times in that moving frame.

Is that logic correct? If so, this would justify the equation l₀ = l/gamma

l = length in moving frame S'
l₀ = length in resting frame S

Yes, exactly so. Going the other way, if we measure the ends of the rod simultaneously in S' at t₂' = t₁', we similarly find that in S, t₂ ≠ t₁. This matters because the rod is not stationary in S.

 

[AdVen]

One end of the object is at x'₁ x'₂. Then, the length of the object in the primed reference frame is L' = x'₂ - x'₁.

Two questions:

Question 1:

What do you mean exactly by the sentence: "One end of the object is at x'₁ x'₂."

Question 2:

Should it not be the case, that x'₂ > x'₁?

PS. I followed the same approach as you did in the thread: Derivation of Proper Time and Proper Space.

 

[stevmg]

Two questions:

Question 1:

What do you mean exactly by the sentence: "One end of the object is at x'₁ x'₂."

Should read one end of the object is at x'₁ and the further end is at x'₂

Question 2:

Should it not be the case, that x'₂ > x'₁?

Yes, that is correct.

PS. I followed the same approach as you did in the thread: Derivation of Proper Time and Proper Space.

Where have you been? I wrote that weeks ago! Not harassing you - just asking.

 

[AdVen]

The math that you wrote below contradicts your above statement.





When you write x'₂ - x'₁ it means that you are measuring in the primed frame.



One more time, L' = γL is wrong. If you did things correctly you should have gotten L' = L/γ

If L' = L/γ then one would expect L = γ L'.

 

[JesseM]

If L' = L/γ then one would expect L = γ L'.

How are you defining L' and L? In post #62 pc2-brazil defined L' as the length of the object in its own rest frame and L as its length in the frame that sees it moving, and likewise in post #1 stevmg said "Suppose I have a thin rod moving along with the primed reference frame" indicating that L' was the rest length of the rod. If L' is the rest length and L is the length in the frame where the rod is moving, the correct equations are L = L'/γ and L' = γL, not the equations you write above. I don't know why starthaus said "One more time, L' = γL is wrong" in post #12, perhaps he/she was confused and thought L was supposed to be the rest length (this is the more common convention when the length contraction equation is written in textbooks and such).

 

[stevmg]

If L' = L/γ then one would expect L = γ L'.

It's been a long time since this topic was started so I'm turned around.

As I understand it, the original "measurement" was L' in the O' (the so-called "moving" frame of reference) and we are after L in O the so-called "stationary" frame of reference.

As starthaus has pointed out, we were to measure L in O such that t₁ = t₂ so that we would get a meaningful length L = x₂ -x₁ where both ends were measured at the same time in O.

In O', where L' was originally situated, since the so-called rod of length L' = x'₂ - x'₁ and since the rod is stationary with respect to O' it doesn't make any difference when t'₁ or t'₂ are because the measurements of x'₁ and x'₂ never change no matter when you measure them in O'

Thus L = L'/\gamma (the rod measures "shorter" in O than in O')

Use these assumptions that I just described to come up with the L = L'/\gamma.

The string is too damn long to go through to find all the mistakes (albeit mine) that were made. Just look at it from scratch:

O' moving at v with respect to O. L' (i.e., x'₁ and x'₂)measured in O' as a given. L' is stationary with respect to O' (thus x'₁ and and x'₂ never, never change). From that we are to calculate L = x₁ - x₂ in O where x₁ and x₂ are measured at the same time (t₁ = t₂).

Again, don't try to fix any prior entries from me. Just go from today post #72 as this has been too long a post and too convoluted (again, probably my fault) to try to untangle.

 

[pc2-brazil]

AdVen: I confirm stevmg's correction to my phrasing. I wrote that phrase in a hurry and ended up forgetting to complete it.
Everyone:
When starthaus said that L' = γL is wrong, I believe he was referring to the case in which L is the length of an object at rest in frame O (posts #62 to #65 conclude the divergence, since I clear up the fact that the situation I was referring to is one in which L' is the rest length).

 

[AdVen]

I am very, very delighted with your response. I can tell you, that my problems with relativistic theory are not primarily the mathematics of it, but much more the 'what is what' of things or 'what is relative to what' or 'which observer in which reference system sees what in another or the same reference sysyem'. I do not know how to tell it differently. I am not a native speaker. You are very clear in this respect for which I am very thankful. You wrote:

I don't know why starthaus said "One more time, L' = γL is wrong" in post #12, perhaps he/she was confused and thought L was supposed to be the rest length (this is the more common convention when the length contraction equation is written in textbooks and such).

I prefer the more common convention and am going to write the derivations and will inform you when I am ready.

 

[AdVen]

How are you defining L' and L? In post #62 pc2-brazil defined L' as the length of the object in its own rest frame and L as its length in the frame that sees it moving, and likewise in post #1 stevmg said "Suppose I have a thin rod moving along with the primed reference frame" indicating that L' was the rest length of the rod. If L' is the rest length and L is the length in the frame where the rod is moving, the correct equations are L = L'/γ and L' = γL, not the equations you write above. I don't know why starthaus said "One more time, L' = γL is wrong" in post #12, perhaps he/she was confused and thought L was supposed to be the rest length (this is the more common convention when the length contraction equation is written in textbooks and such).

I hope I do understand your words correctly if I conclude as follows:

If L is the rest length and L' is the length in the frame where the rod is moving, the correct equations are L' = L/γ and L = γL'.

 

[AdVen]

Hi JesseM,

I have made a mathematical derivation of the Length Contraction (Proper Length) now by following the more common convention. Please, would you check whether it is finally oké now. You can find the derivation by clicking underneath:

http://www.socsci.ru.nl/~advdv/ProperLengthFinal.pdf

I really hope it is oké now and I would appreciate it very much if you would look into it.

Yours, AdVen.

 

[stevmg]

Hi JesseM,

I have made a mathematical derivation of the Length Contraction (Proper Length) now by following the more common convention. Please, would you check whether it is finally oké now. You can find the derivation by clicking underneath:

http://www.socsci.ru.nl/~advdv/ProperLengthFinal.pdf

I really hope it is oké now and I would appreciate it very much if you would look into it.

Yours, AdVen.

Nice article AdVen. In my opinion, which means nothing on this forum as I am, too, a complete novice, if the given length L is in the rest frame O, the "measured" length L' in S' (the moving frame) at t'₁ = t'₂ is L' = L/\gamma.

I hope JesseM confirms this which will make my understanding likewise correct.

 

[JesseM]

Hi JesseM,

I have made a mathematical derivation of the Length Contraction (Proper Length) now by following the more common convention. Please, would you check whether it is finally oké now. You can find the derivation by clicking underneath:

http://www.socsci.ru.nl/~advdv/ProperLengthFinal.pdf

I really hope it is oké now and I would appreciate it very much if you would look into it.

Yours, AdVen.

Yup, the derivation looks good to me.

I hope I do understand your words correctly if I conclude as follows:

If L is the rest length and L' is the length in the frame where the rod is moving, the correct equations are L' = L/γ and L = γL'.

Right.

 

[AdVen]

Yup, the derivation looks good to me.

Right.

I suppose that 'yup' means 'yes'. Anyhow, I am very grateful that you have looked into it.

I am now trying to derive in a similar way the Time Dilatation Proper Time formula. I hope you will soon hear from me. If you might have any suggestions, please, let me hear.

 

[starthaus]

I suppose that 'yup' means 'yes'. Anyhow, I am very grateful that you have looked into it.

You can shorten your proof considerably this way:

x'=\gamma(x-vt)
t'=\gamma(t-xv/c^2

Differentiating the above you obtain:dx'=\gamma(dx-vdt)
dt'=\gamma(dt-vdx/c^2)

Now:

L'=dx' is the length you measure in the moving frame S' and L=dx is the length of the rod measured in the co-moving frame S (proper length). In order for your measurement in S' to be valid, you must mark both ends simultaneously in S':

dt'=0

so:

0=\gamma(dt-vdx/c^2)

meaning that :

dt=vdx/c^2=vL/c^2

Therefore:

L'=dx'=\gamma(dx-vdt)=\gamma(L-v^2L/c^2)=L/\gamma

The derivation for time dilation follows a similar pattern:

dx=0 (the events a happen at the same location in frame S)

dt'=\gamma dt

That's it.

 

[JesseM]

I am now trying to derive in a similar way the Time Dilatation Proper Time formula. I hope you will soon hear from me. If you might have any suggestions, please, let me hear.

With the time dilation equation, just remember that the equation deals specifically with the case of two events that happen at the same spatial location in one of the two frames, but different times (like successive readings on a clock that's at rest in the unprimed frame). This makes it pretty simple, since in the equation dt' = gamma*(dt - vdx/c^2) [or if you prefer, the equation (t'₂ - t'₁) = gamma*((t₂ - t₁) - v*(x₂ - x₁)/c^2)], you know dx = 0.

 

[AdVen]

Thanks a lot. I am very glad with your suggestion. I am going to look into it tomorrow. I hope to reply tomorrow.

 

[AdVen]

With the time dilation equation, just remember that the equation deals specifically with the case of two events that happen at the same spatial location in one of the two frames, but different times (like successive readings on a clock that's at rest in the unprimed frame). This makes it pretty simple, since in the equation dt' = gamma*(dt - vdx/c^2) [or if you prefer, the equation (t'₂ - t'₁) = gamma*((t₂ - t₁) - v*(x₂ - x₁)/c^2)], you know dx = 0.

I already knew what you are telling in the first sentence. However, thanks a lot for this idea too. I am also very glad with this suggestion. I am going to look into it tomorrow also. I hope to reply tomorrow.

 

[AdVen]

You need to calculate x₂ - x₁ for the condition: t₂ = t₁. You are inadvertently calculating it for t'₂ = t'₁, this is why you get the error.

I am trying to derive the formula for the time dilatation. I consider the clock is situated in the rest frame S and the observer is situated in the moving frame S'. I want to derive the length of time on the clock as seen by the observer in the moving frame. Therefore I think I need to calculate t'₂-t'₁ for the condition: x'₂ = x'₁. This leads to x₂-x₁ = (t₂-t₁)*v. As a final result I get t'₂-t'₁ = (t₂-t₁) * gamma, where gamma is the Lorentz factor.

Is this correct? I am afraid it is not. At least in Wikipedia I get a different result: http://en.wikipedia.org/wiki/Time_dilation

 

[starthaus]

I am trying to derive the formula for the time dilatation. I consider the clock is situated in the rest frame S and the observer is situated in the moving frame S'. I want to derive the length of time on the clock as seen by the observer in the moving frame. Therefore I think I need to calculate t'₂-t'₁ for the condition: x'₂ = x'₁. This leads to x₂-x₁ = (t₂-t₁)*v. As a final result I get t'₂-t'₁ = (t₂-t₁) * gamma, where gamma is the Lorentz factor.

Is this correct? I am afraid it is not. At least in Wikipedia I get a different result: http://en.wikipedia.org/wiki/Time_dilation

It's correct, you are doing fine. For a simpler derivation, see the last two lines in post 80.

 

[JesseM]

I am trying to derive the formula for the time dilatation. I consider the clock is situated in the rest frame S and the observer is situated in the moving frame S'. I want to derive the length of time on the clock as seen by the observer in the moving frame. Therefore I think I need to calculate t'₂-t'₁ for the condition: x'₂ = x'₁.

If the clock is at rest in the unprimed frame, then the condition should be x₂ = x₁ (the clock's unprimed position coordinate remains unchanged), not x'₂ = x'₁.

This leads to x₂-x₁ = (t₂-t₁)*v.

If you want to start from the part of the Lorentz transformation that deals with x-coordinates, you'd use (x₂-x₁) = gamma*((x'₂-x'₁) + v*(t'₂-t'₁)), and then with x₂-x₁ = 0 you're left with x'₂-x'₁ = -v*(t'₂-t'₁). However, if you're trying to derive the time dilation equation it's much better to start with the part of the Lorentz transformation that deals with t-coordinates, namely (t'₂ - t'₁) = gamma*((t₂ - t₁) - v*(x₂ - x₁)/c^2), then if you substitute in (x₂ - x₁) = 0 you're left with (t'₂ - t'₁) = gamma*(t₂ - t₁), which is the correct time dilation equation.

As a final result I get t'₂-t'₁ = (t₂-t₁) * gamma, where gamma is the Lorentz factor.

Oddly this equation is correct even though your earlier step x'₂ = x'₁ was incorrect! Maybe you made an algebra error too?

Is this correct? I am afraid it is not. At least in Wikipedia I get a different result: http://en.wikipedia.org/wiki/Time_dilation

Why do you think wikipedia gives a different result? The time dilation formula in this section is the same as the one you just wrote.

 

[AdVen]

Oddly this equation is correct even though your earlier step x'₂ = x'₁ was incorrect! Maybe you made an algebra error too?

Why do you think wikipedia gives a different result? The time dilation formula in this section is the same as the one you just wrote.

You can find the mathematics (without text) on:

http://www.socsci.ru.nl/~advdv/TimeDilatationFinal.pdf

Hopefully, you can understand. Otherwise I will add explaining text to it.

 

[JesseM]

You can find the mathematics (without text) on:

http://www.socsci.ru.nl/~advdv/TimeDilatationFinal.pdf

Hopefully, you can understand. Otherwise I will add explaining text to it.

OK, the steps of going from lines 2 and 3 to line 4 indicate you are assuming x'₁ = x'₂, which as I said is the wrong physical assumption if you want the coordinates to be those of a clock at rest in the unprimed frame. Still, the assumption would work if they were the coordinates of a clock at rest in the primed frame, in which case the correct final time dilation equation should be t'₂ - t'₁ = (t₂ - t₁)/gamma, which is what you got in the pdf. In the earlier post you had said "As a final result I get t'₂-t'₁ = (t₂-t₁) * gamma" which is why I was confused (since here you are multiplying the unprimed time difference by gamma rather than dividing it by gamma), but that's not the equation that appears in the pdf.

 

[AdVen]

OK, the steps of going from lines 2 and 3 to line 4 indicate you are assuming x'₁ = x'₂, which as I said is the wrong physical assumption if you want the coordinates to be those of a clock at rest in the unprimed frame. Still, the assumption would work if they were the coordinates of a clock at rest in the primed frame, in which case the correct final time dilation equation should be t'₂ - t'₁ = (t₂ - t₁)/gamma, which is what you got in the pdf. In the earlier post you had said "As a final result I get t'₂-t'₁ = (t₂-t₁) * gamma" which is why I was confused (since here you are multiplying the unprimed time difference by gamma rather than dividing it by gamma), but that's not the equation that appears in the pdf.

I am very, very grateful for your comment. I am going to read this very carefully and will reply as soon as possible. I think my major problem is understanding what is what (I do not know how to say it differently, as I am not a native speaker). You have many possibilities:

the clock at rest in the unprimed frame,
the clock at rest in the primed frame,
the observer at rest in the unprimed frame,
the observer at rest in the primed frame

and so on

and how these are related to the different assumptions:

x₁ = x₂
x'₁ = x'₂

Since the time I am studying special relativity I have great diffculties with what I call above 'what is what'.

Among other things the expression the 'moving observer' gives me difficulties. I know he/she is moving with respect to the clock. But the usual Lorentz transformation is about a rest frame S and a moving frame S'. Now, if the clock is located in S' then the so called 'moving observer' is located in the rest frame, which is not moving. I hope you can understand my confusion.

 

[AdVen]

Still, the assumption would work if they were the coordinates of a clock at rest in the primed frame, in which case the correct final time dilation equation should be t'₂ - t'₁ = (t₂ - t₁)/gamma, which is what you got in the pdf.

Yes, I imagine the following situation:

S: frame at rest (x and t are coordinates in S).

S': moving frame with respect to S (x' and t' are coordinates in S').

Lorentz transformations::

x' = γ(x - vt)

t' = γ(t - vx/c²)

Where γ is the Lorentz factor and c is the speed of light in vacuum.

Clock: clock is at rest in S'.

t₂ - t₁: time difference observed by the observer, which is located in S, when looking at the clock in S'. This observer is the so-called moving observer. Although he is NOT moving with respect to S, he IS moving relative to S' and, therefore also relative to the clock in S', which is at rest in S'.

t'₂ - t'₁: time difference observed by the observer at rest in S' when looking at the clock in S' . This observer is located in S'.

The expression t'₂ - t'₁ = (t₂ - t₁)/γ expresses the fact that for the resting observer the period of the clock is shorter then for the moving observer.

Do you think, that this correct now?

Thanks a lot for your concern and for your time and effort.